資料結構C++版-棧

標籤：std   匹配   bin   ase   nat   等於   alt   else   ace   

 

一、概念

 

 

二、應用執行個體

1.進位轉換

#include <stdlib.h>#include <iostream>#include <string>#include "MyStack.h"#include "Coordinate.h"using namespace std;#define  BINARY 2#define  OCTONARY 8#define  HEXADECIMAL 16void main ( ){     //MyStack<int> s(30);    MyStack<char> s(30);    char num[]="0123456789ABCDEF";    int N=1348;    int mod=0;    while (N!=0)    {        mod=N%16;        s.push(num[mod]);        N=N/16;    }    s.stackTraverse(false);        /*        int elem=0;        for (int i=s.stackLength()-1;i>=0;i--)    {        s.pop(elem);        cout<<num[elem];    }*/    /*while(!s.stackEmpty())    {        s.pop(elem);        cout<<num[elem];    }*/    system("pause");    } 

 

2.括弧匹配

MyStack<char> s(30);   //存放未匹配的括弧    MyStack<char> s1(30);  //存放棧頂括弧的另一半    char str[] = "([])[";    //存放待匹配文本目標，要求無空格 e.g.   [()()]  ([]([]))    char current=0;  //當前括弧需要匹配的另一半    for (int i=0;i<strlen(str);i++)    {        if (current!=str[i])        {            s.push(str[i]);            switch(str[i])            {            case  ‘[‘:    //case 後面資料類型是int，單個字元會轉換成其ASC碼                if (current!=0)                {                    s1.push(current);                }                current=‘]‘;                break;            case ‘(‘:                if (current!=0)                {                    s1.push(current);                }                current=‘)‘;                break;            default:                cout<<"括弧不匹配."<<endl;  //default後面語句可以注釋掉，因為current不等於str[i]時str[i]就會入棧，第一個棧不為零匹配就會失敗            }                    }         else        {            char elem;            s.pop(elem);            if(!s1.pop(current))            {                current=0;            }        }    }if (s.stackEmpty())    {        cout<<"括弧匹配"<<endl;    }     else    {        cout<<"括弧不匹配"<<endl;    }    system("pause");

 

 

 

資料結構C++版-棧